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Finding inflection points

  1. May 12, 2007 #1

    ssb

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    1. The problem statement, all variables and given/known data

    For many differential equations, the easiest way to find inflection points is to use the differential equation rather than the solution itself. To do this, we can compute [tex]y''[/tex] by differentiating [tex]y'[/tex], remembering to use the chain rule wherever [tex]y[/tex] occurs. Next, we can substitute for [tex]y'[/tex] by using the differential equation and setting [tex]y' = 0[/tex]. Then we can solve for [tex]y[/tex] to find the inflection points. (Keep in mind here that solving for [tex]y[/tex] can also produce some equilibrium solutions, which may not be inflection points!)

    Use the technique described above to find the inflection point for the solutions of the differential equation


    [tex]y'=r(1-\frac{y}{L})y[/tex]

    your answer may contain [tex]L[/tex] and [tex]r[/tex]



    [tex]y = ?[/tex]





    3. The attempt at a solution


    I differentiated the given equation and set it equal to zero, then I solved it for y. My answer was Lr/4 but this is wrong according to webworks.

    The equation I got when I differentiated [tex]y'=r(1-\frac{y}{L})y[/tex] was [tex]y'' = r-((4y)/L)[/tex]

    i know the answer is [tex]L/2[/tex] but I dont know how to get there.
     
    Last edited: May 12, 2007
  2. jcsd
  3. May 12, 2007 #2
    y' = ry(1 - y/L)
    distribute ...
    y' = ry - ry2/L
    differentiate ...
    y" = r - 2ry/L
    set y" = 0 ...
    r - 2ry/L = 0
    r(1 - 2y/L) = 0
    1 = 2y/L
    y = L/2
     
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